Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		double[] arr = new double[n + 1];
		for (int i = 1; i <= n; i++)
			arr[i] = s.nextDouble();
		double res = 0.0;
		double[][] dp=new double[n+1][n+1];
		boolean[][] vis=new boolean[n+1][n+1];
		for (int head = 0; head <= n; head++) {
			int tail = n - head;
			if (head > tail)
				res += solve(arr, n, head,dp,vis);
		}
		System.out.println(res);
	}

	public static double solve(double[] arr, int n, int k,double[][] dp, boolean[][] vis) {
		if (k < 0)
			return 0.0;
		if (n == 0)
			return (k == 0) ? 1.0 : 0.0;
		if(vis[n][k])
			return dp[n][k];
		else {
			vis[n][k]=true;
			dp[n][k]=arr[n] * solve(arr, n - 1, k - 1,dp,vis) + (1.0 - arr[n]) * solve(arr, n - 1, k,dp,vis);
		return dp[n][k];
		}
	}

}